The last stage in the fusion of hydrogen to form helium in the Sun involves the formation of an alpha particle (4He) from two 3He nuclei:
3He + 3He 4He + p + p
The binding energy for 3He is 2.57 MeV per nucleon and that for 4He is
7.07 MeV per nucleon. Calculate the amount of energy released in this reaction.
Is 12.9 MeV close? Thanks
To calculate the amount of energy released in the fusion reaction, we need to calculate the difference in binding energies between the reactants (two 3He nuclei) and the product (4He nucleus) and then convert it to MeV.
The total binding energy of the two 3He nuclei is:
2 * (2.57 MeV/nucleon) * 3 nucleons = 15.42 MeV
The binding energy of the 4He nucleus is:
(7.07 MeV/nucleon) * 4 nucleons = 28.28 MeV
The energy released in the reaction is the difference between the initial and final binding energies:
Energy released = Initial binding energy - Final binding energy
Energy released = 28.28 MeV - 15.42 MeV
Energy released = 12.86 MeV
So the amount of energy released in this reaction is approximately 12.86 MeV, which is very close to the value you mentioned as 12.9 MeV.